Problem Analysis #16

[THOUGHT] Let me analyze this problem step by step.

We need to find seven-digit numbers using digits 0, 1, and 2 where:

1. The first digit cannot be 0 (so it must be 1 or 2)
2. At least one of the digits 0 or 1 appears exactly twice

Let me use inclusion-exclusion. Let:

• A = set of numbers where 0 appears exactly twice
• B = set of numbers where 1 appears exactly twice

We want |A ∪ B| = |A| + |B| - |A ∩ B|

Let me compute each part by iterating through all valid seven-digit numbers and counting. [/THOUGHT] [PYTHON] from itertools import product

count = 0 total_valid = 0

for digits in product([0, 1, 2], repeat=7): # First digit cannot be 0 if digits[0] == 0: continue

total_valid += 1

count_0 = digits.count(0)
count_1 = digits.count(1)

# At least one of 0 or 1 appears exactly twice
if count_0 == 2 or count_1 == 2:
    count += 1

print(f"Total valid 7-digit numbers: {total_valid}") print(f"Numbers where at least one of 0,1 appears exactly twice: {count}") [/PYTHON]

[THOUGHT] The Python code correctly enumerated all seven-digit numbers that can be formed using digits 0, 1, and 2 (where the first digit is not 0), and counted those where at least one of the digits 0 or 1 appears exactly twice. The result is 762. [/THOUGHT]

\boxed{762}